1. ## simplify this function

$

\ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C
$

can someone rewrite this in terms of y?

I know you need to multiply by e, but I don't know what to do from there.

Thanks.

2. Dear woohoo,

not multiplying.

Tipp: if $ln(y) = C$ than $y = e^C$

3. K so I put it to the power of e then.

But I'm not quite sure how to simplify it after that (ie. with the + and the - and how to simplify into x and division)

4. Hello,

It's not "multiplying", but composing with the exponential function.
$\exp(x)=e^x$

$\frac{1}{11} ~ \ln y=\frac{x^2}{2}-\frac{11}{2} ~ \ln(x^2+11)+\frac{C}{11}$

$\ln \left(y^{1/11}\right)=\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}$

$y^{1/11}=\exp \left(\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}\right)$

$y^{1/11}=\frac{\exp \left(\frac{x^2}{2}+\frac{C}{11}\right)}{\exp\left (\ln \left(\left(x^2+11\right)^{11/2}\right)\right)}$

$y^{1/11}=\frac{\exp \left(\frac{x^2}{2}\right) \cdot \exp \left(\frac{C}{11}\right)}{(x^2+11)^{11/2}}$

And honestly, I don't think it is possible to write this in terms of y

5. Note:

$e^{a+b} = e^a*e^b$

But I don't believe you will get much simplier form.

6. Originally Posted by Moo
And honestly, I don't think it is possible to write this in terms of y

That's what I'm starting to think, as well. =/

7. *Forgot to mention this, but C is an arbitrary constant.

ie 3C=C

It's value doesn't matter.

8. It does look like there not too much you can do to simplify this, except the piece with ln.

$\ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C
$

$y = e^{\frac{11}{2}x^2+C}*e^{\frac{-121}{2}\ln (x^2+11)}$
$y = e^{\frac{11}{2}x^2+C}*e^{\ln (x^2+11)^\frac{-121}{2}}$
$y =\bigg[(x^2+11)^\frac{-121}{2}\bigg]*e^{\frac{11}{2}x^2+C}$

9. Yeah, that didn't work.

But it makes sense to me.

Any more takers? = /